Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))

Q is empty.


QTRS
  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R consists of the following rules:

half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))


Q DP problem:
The TRS P consists of the following rules:

BITS1(s1(x)) -> HALF1(s1(x))
BITS1(s1(x)) -> BITS1(half1(s1(x)))
HALF1(s1(s1(x))) -> HALF1(x)

The TRS R consists of the following rules:

half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

BITS1(s1(x)) -> HALF1(s1(x))
BITS1(s1(x)) -> BITS1(half1(s1(x)))
HALF1(s1(s1(x))) -> HALF1(x)

The TRS R consists of the following rules:

half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
QDP
                ↳ QDPAfsSolverProof
              ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

HALF1(s1(s1(x))) -> HALF1(x)

The TRS R consists of the following rules:

half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

HALF1(s1(s1(x))) -> HALF1(x)
Used argument filtering: HALF1(x1)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
QDP

Q DP problem:
The TRS P consists of the following rules:

BITS1(s1(x)) -> BITS1(half1(s1(x)))

The TRS R consists of the following rules:

half1(0) -> 0
half1(s1(0)) -> 0
half1(s1(s1(x))) -> s1(half1(x))
bits1(0) -> 0
bits1(s1(x)) -> s1(bits1(half1(s1(x))))

The set Q consists of the following terms:

half1(0)
half1(s1(0))
half1(s1(s1(x0)))
bits1(0)
bits1(s1(x0))

We have to consider all minimal (P,Q,R)-chains.